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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Let <span class="process-math">\({\bf x}={\bf T}\,\vec{y}\text{,}\)</span> then the equation becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\vec{y}^{\prime}={\bf T}^{-1} \, {\bf A}\, {\bf T}\,\vec{y}+{\bf T}^{-1}  \, {\bf g}(t),
\end{equation*}
</div>
<p class="continuation">where</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf T}^{-1}=\frac{1}{5}\left(
\begin{array}{cc}
1 &amp; -1\\
4 &amp; 1
\end{array}
\right),\quad
{\bf T}^{-1}  \, {\bf g}(t)=\frac{1}{5}\left(
\begin{array}{c}
e^{-2t}+2 e^t\\
4 e^{-2t}-2e^t
\end{array}
\right).
\end{equation*}
</div>
<p class="continuation">In component form, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{array}{c}
y_1^{\prime}=-3y_1+\frac{1}{5} (e^{-2t}+2 e^t),\\
y_2^{\prime}=2y_2+\frac{1}{5} (4 e^{-2t}-2 e^t),
\end{array}
\end{equation*}
</div>
<p class="continuation">which lead to solutions</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{array}{c}
y_1=C_1 e^{-3t}+\frac{1}{5}(e^{-2t}+e^t/2 )\\
y_2=C_2 e^{2t}+\frac{1}{5}(2 e^t-e^{-2t}).
\end{array}
\end{equation*}
</div>
<p class="continuation">Then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}={\bf T}\,\vec{y}=C_1 \left(
\begin{array}{c}
1\\
-4
\end{array}
\right) e^{-3t}+C_2 \left(
\begin{array}{c}
1\\
1
\end{array}
\right) e^{2t}+ \left(
\begin{array}{c}
0\\
-1
\end{array}
\right) e^{-2t}+ \left(
\begin{array}{c}
1/2\\
0
\end{array}
\right) e^{t}.
\end{equation*}
</div>
<span class="incontext"><a href="sec6_6.html#p-295" class="internal">in-context</a></span>
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